How do I identify the x-intercept(s) and vertical asymptote(s): #y=(x^3+27)/(3x^2+x)#?
1 Answer
Feb 26, 2015
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To identify the x-intercepts, you want to ask yourself: "where does the graph hit the x-axis, aka: what is x when y=0?"
So let y = 0 and solve for x:
#0=(x^3+27)/(3x^2+x)#
In order for this fraction to equal 0, the numerator of the fraction must equal 0 (remember: denominator = 0 -> undefined)
#0=(x^3+27)#
#x^3=-27#
#x=-3#
So the x-intercept: (-3,0) -
To identify the vertical asymptotes, we first try and simplify the function as much as possible and then look at where it is undefined
#y=(x^3+27)/(3x^2+x)# is already simplified
Undefined when denominator = 0:#(3x^2+x)=0#
#x(3x+1)=0#
#x=0, 3x+1=0#
Vertical asymptotes:#x=0, x=-1/3#
